NICOLAS-SERRE THEORY
Let $F \in Z/2[[x]]$ be $x+x^9+x^{25}+...$, the exponents being the odd squares, and $V$ be the space spanned by the $F^k$ with $k$ odd. Nicolas and Serre define formal Hecke operators $T_n$, $n$ odd, on $Z/2[[x]]$ and show:
1) $V$ is stable under the $T_n$. If $n$ is not a square, then $T_{n}(F^{k})$, $k$ odd, is a sum of various $F^j$ with $j$ odd and smaller than $k$. (This comes from modular form theory for the full modular group).
2) The only elements of $V$ annihilated by both $T_3$ and $T_5$ are $0$ and $F$.
3) There are $m_{a,b}$ forming a basis of $V$ such that $m_{0,0}$ is $F$, the $m_{a,b}$ with $a$ and $b$ not both $0$ are divisible by $x^2$, and $T_3$ and $T_5$ reduce $a$ by 1 and $b$ by 1 respectively.
4) In the action of the algebra spanned by the $T_n$, each $T_n$ "acts as a power series in $T_3$ and $T_5$". This tells us that a certain completion of this Hecke algebra is a $2$-variable power series ring over $Z/2$ in $T_3$ and $T_5$.
The proofs of (2)-(4) are rather technical and rely on a certain "code".
CONJECTURAL ANALOGS
I've experimentally found some analogs to the above related to modular forms of level $N$ when $N$ is 3, 5, 7, or 11. The situation when $N=3$ is particularly nice. Explicitly let $D \in Z/2[[x]]$ be $x+x^{25}+x^{49}+...$, the exponents being the squares prime to $6$, and let $V$(plus) be spanned by the $D^k$ with $k \equiv 1 (\text{mod}\ 6)$. Then:
1*) V(plus) is stable under the $T_n$ with $n \equiv 1(\text{mod}\ 6)$. If $n$ is a non-square and $\equiv 1 (\text{mod}\ 6)$ then $T_n(D^k)$ is a sum of $D^j$ for various $j$ with $j\equiv 1 (\text{mod}\ 6)$ and $j$ smaller than $k$. (This is a theorem, whose proof uses modular forms of level 3).
2*) I believe that the only elements of V(plus) annihilated by $T_7$ and $T_{13}$ are $0$ and $D$.
3*)I believe there are $m_{a,b}$ precisely as in 3) with $F, T_3$ and $T_5$ replaced by $D, T_7$, and $T_{13}$. (I've calculated these when $a+b$ is at most 6. For example, $m_{4,2}$ is $(D^{73})+(D^{121})+(D^{145})$).
4*)I believe that in the action of the algebra spanned by the $T_n$ with $n\equiv 1 (\text{mod}\ 6)$ on V(plus) each such $T_n$ acts as a formal power series in $T_7$ and $T_{13}$. Also if $n$ is 1 mod 24, then $T_n$ acts as a power series in $(T_7)^2$ and $(T_{13})^2$, and in particular $(T_1)+(T_{25})=(T_5)^2$ acts by $(T_7)^2+(T_{13})^2$+(higher order terms). (This has implications for the structure of a completion of the algebra spanned by the $T_n$ with $(n,6)=1$ on the space spanned by the D^k with $(k,6)=1)$.
QUESTION: posed to those who understand the Nicolas-Serre code. Can one use some modification of the code to establish the truth of my beliefs stated above? Any thoughts about proving these conjectures would be welcome.
Remark 1: When $N=5$ there is a rather more complicated analog of V(plus) on which the T_n with n=1 or 3 mod 8 act. Results like (1) can again be proved. Results like (2)-(4) seem to hold with $T_3$ and $T_5$ replaced by $T_3$ and $T_{11}$. And if $n$ is 1 or 9 mod 40 then $T_n$ seems to be a power series in $(T_3)^2$ and $(T_{11})^2$. In particular $(T_7)^2$ acts by $(T_3)^2+(T_{11})^2$+(higher order terms). Again there would be implications for the structure of a completed Hecke algebra.
Remark 2: When $N=3$ there are similar apparent results for V(minus), the space spanned by the $D^k$ with k=5 mod 6. And when $N=5$, there is an analog of this V(minus) with similar apparent results.
EDIT___A GENERAL FRAMEWORK
I'll place the calculations I've made into a general setting, and then explain how the case N=3 described under CONJECTURAL ANALOGS fits into this expanded framework.
Let N be an odd prime and HE the "level N shallow Hecke algebra" spanned by the T_k: Z/2[[x]]-->Z/2[[x]] with(k,2N)=1. Let F be as above and G be F(x^N). I'll adopt the notation of my question 138495--"Are these two subspaces of Z/2[[x]] the same?" In particular, M is the integral closure of Z/2[G] in Z/2(F,G) viewed as a subspace of Z/2[[x]]. M has a modular forms interpretation which shows that it (and its subspaces M(odd) and C) are stable under HE. Indeed each of these is an increasing union of finite dimensional HE stable subspaces. Each of these finite dimensional subspaces is evidently annihilated by a product of powers of maximal ideals of HE. It's known, I believe, that only finitely many of these maximal ideals appear altogether. So in particular, C is a direct sum of finitely many HE stable subspaces, C(J), with the maximal ideal J acting locally nilpotently on C(J).(One case of interest is when J=Ann(F), the annihilator of F in HE. This ideal contains T_p for all primes p other than 2 or N. When N=3,5, or 7, I'm informed that Ann(F) is the only maximal ideal that appears, so that C=C(J).)
Now let pr:Z/2[[x]]-->Z/2[[x]] be the map which removes from each h in Z/2[[x]] all terms in which N divides the exponent. Then pr(C(J)) is HE stable with J acting locally nilpotently on it. As in Nicolas-Serre one can then construct a "J-completion" of HE acting faithfully on pr(C(J)). A general question is:
WHAT IS THE STRUCTURE OF THIS J-COMPLETION?
The hope that, as in Nicolas-Serre, the J-completion is a 2 variable power series ring proves mistaken. But in some cases the following seems to hold:
There are one or more index 2 subgroups of Z/(8N)* such that when one replaces HE by the subalgebra HE# spanned by the T_k with k in one of these subgroups, and C(J) by the subspace C(J)# consisting of those h in C(J) for which all the exponents that appear are either divisible by N or lie in the subgroup, then the resulting J-completion of HE# is a 2 variable power series ring. Furthermore, in a number of these cases, the J-completion of HE appears to be isomorphic to the non-reduced ring Z/2[[x,y,z]]/(z^2).
In the examples I've looked at, N=3,5,7 and 11. For each of these N, C is the set of elements of M(odd) whose trace from Z/2(F,G) to Z/2(G) is 0, and there is a Z/2[G^2] basis Ck of C with k odd, k between 0 and 2N, given in question 138495 with nice properties. Set Dk equal to pr(Ck), and extend the definition of Dm to all odd m so that whenever m=k+2N, then Dm=(G^2)Dk. Then Dm=0 when N divides m, while the remaining Dm are a Z/2 basis of pr(C). Calculations with this basis are very convenient. I won't go into the details of the calculations now, but I'll explain how the case N=3 under CONJECTURAL ANALOGS fits into this new setting.
Suppose then that N=3 and J=Ann(F). Look at the N=3 paragraph under SOME REMARKABLE FACTS in question 138495. Since C1=F, D1=pr(F) which is the D described above. Then if r=6k+1, Dr=(G^2k)*D. But classical results show that G=D^3, so that Dr=D^r for all r that are 1 mod 6. Similarly, C5=(F^2)G, so that D5=pr(F^2)G=(D^2)G=D^5, and it follows that Dr=D^r for all r that are 5 mod 6. Thus pr(C) is just the space V spanned by the D^r with (r,6)=1.
Now as I've noted, C=C(J) in this setting. Now {1,7,13,19} is an index 2 subgroup of (Z/24)*, and a Z/2 basis of the corresponding subgroup pr(C#) of pr(C) is given by the D^k with k=1 mod 6. It follows that pr(C(J)#) is just the space V(plus) spanned by these D^k, described under CONJECTURAL ANALOGS, and we are in precisely the situation given there.
EDIT(Sept. 1)__WHAT THE COMPUTER SUGGESTS IN LEVEL 11
The 1-dimensional subspace {0,F} of C is always HE stable. When N=11 there is another HE stable 1-dimensional subspace {0,t} of C, with t as in the N=11 paragraph of my question 138495. (In fact, t=C1+C3+C5+C9+C15). One way to see HE stability is the following--t^12=FG, the reduction of the expansion of the modular form delta(z)delta(11z). So t is the reduction of the expansion of (eta(z)eta(11z))^2, and this last is the weight 2 newform for Gamma_0 (11). Now Ann(t) and Ann(F) are maximal ideals in HE. Write C(t) and C(F) for C(Ann(t)) and C(Ann(F)). I've been informed that C is the direct sum of C(t) and C(F). Here's what the computer suggests for the Ann(t)-completion of HE acting on pr(C(t)).
(*) The above completion is a 2-variable power series ring over Z/2, with an element of square 0 adjoined. More precisely the map from Z/2[[X,Y,Z]] to the Ann(t)-completion that sends X,Y and Z to T_3 +I, T_5 +I and T_7 is onto and the kernel is generated by f^2 where f=Z+X+X^2+XY+Y^2+(X^2)Y+X(Y^2)+Y^3+higher degree stuff in X and Y.
Remarks__See my question 137260--Questions(related to deformation theory?)... for what the above tells us about a very natural attempted generalization of Nicolas-Serre to level 11. I have conjectures similar to (*), supported by the computer, for the Ann(F)-completion of HE acting on pr(C(F)) when N=3,5 or 11. But N=7 is more complicated; I don't understand it at this time.
Assume now that N=11. Since T_3(t)=t, T_3+ I acts nilpotently on C(t), while T_3 acts nilpotently on C(F). So we can use the calculations made of the various T_3(Dm) as sums of various Dk's to decompose each Dm into its pr(C(t)) and pr(C(F))-components for a wide range of m, and then calculate the effect of T_3, T_5 and T_7 on each of the pr(C(t))-components.
I now make use of two index 2 subgroups of (Z/88)*, G1 and G2. The first consists of the k that are squares mod 11, and the second of the 4n+1 that are squares mod 11 and the 4n+3 that are non-squares. Use G1 and G2 to define subalgebras HE(1#) and HE(2#) of HE, as well as subspaces C(t,1#) and C(t,2#) of C(t). Using the calculations outlined in the last paragraph, I empirically find:
(A)--If we replace F by pr(t)=[1,3,5,9,15] (this is shorthand for D1+D3+D5+D9+D15), T3 and T5 by T3 +I and T_5 +I, the algebra spanned by the T_n by HE(1#), and V by pr(C(t,1#)), then 2),3) and 4) under NICOLAS-SERRE THEORY still hold.
(B)--If we replace F by [1,5,9], T3 and T5 by T5 +I and T7, the algebra spanned by the T_n by HE(2#), and V by pr(C(t,2#)), then 2),3) and 4) under NICOLAS-SERRE THEORY still hold. (I've calculated many m_(a,b) both for A and for B. For example in A, m_(1,7)=[23,31,47,71,191,223]).
Now let x,y and z be the images of T3 +I, T5 +I and T7 in the Ann(t)-completion of HE, acting on pr(C(t)). From A and B one should be able to show:
1) z^2 is a power series in x^2 and y^2. (First one shows that this is true with HE replaced by HE(1#) and C(t) by C(t,1#)).
2) More precisely f^2=0, where f=z+x+x^2+xy+y^2+(x^2)y+x(y^2)+y^3+higher degree stuff in x and y. (For this one needs to write z^2 as a power series in x^2 and y^2 to the needed accuracy. This is accomplished by seeing what (T_7)^2 does to m_(7,1), m_(5,3), m(3,5) and m(1,7)).
3) Every element of the J-completion of HE acting on pr(C(t) is a power series in x,y and z.(For the elements of the J-completion of HE(1#)(resp. HE(2#)) acting on C(t) are power series in x and y(resp. y and z). And the two subalgebras generate HE.
Putting 2) and 3) together should give the desired (conjectural) structure theorem for our Ann(t)-completion of HE.
EDIT(Sept. 10) WHAT THE COMPUTER SUGGESTS IN LEVELS 3 and 5
When N=3 or 5, J can only be Ann(F), and so C=C(J). The computer indicates that in each case there is a triple of integers {n_1,n_2,n_3} with the following properties:
I.--Let a and b be two elements of the triple and G be the index 2 subgroup of (Z/8N)* generated by a,b and the squares. Use G to define the subalgebra HE# of HE and the subspace C# of C, as under WHAT IS THE STRUCTURE OF THIS J-COMPLETION? Then 2),3) and 4) under the heading NICOLAS-SERRE THEORY hold provided we replace:
The space spanned by the T_n by HE#, V by C#, T_3 by T_a and T_5 by T_b.
In particular, the Ann(F)-completion of HE# acting on C# is a power series ring in T_a and T_b over Z/2.
II.--The Ann(F)-completion of HE acting on C is a power series ring in x=T_(n_1), y=T_(n_2) and z =T_(n_3) with a single relation f^2=0, where f=x+y+z+(higher degree stuff).
IIa.--When N=3, n_1=5, n_2=7, n_3=13 and f=x+g where g=y+z+y^3+y*z^2+z^3+(higher degree stuff in y and z.
IIb.--When N=5, n_1=3, n_2=7, n_3=11 and f=y+g where g=x+z+x^3+x*z^2+z^3+(higher degree stuff in x and z.
In support of (I), I calculated the m_(i,j) for each HE# and C# whenever i+j is 6 or less. If (I) holds, an argument similar to the one I made in the last edit, should give the remaining results.
Remark__I think there are similar results when N=11, J=Ann(F), though I haven't carried out the calculations of the m_i,j very far. But N=7 is, as I've noted, very much different.